Give you set of meetings start time and end time, count how many meeting rooms needed.
For example:
[[2, 5], [1, 3], [2, 7], [6, 8]] // needs 3 meeting rooms
We can use a memo table to store the at which time how many rooms needed. It is sort of like Knapsack problem. Check out this post for similar question.
Code:
function* generateAry(start, end) { let i = start; while (i <= end) { yield i; i++; } } // O(N*K): K: number of hours, N: number of room // S(K) function availableRoom(arys) {const { min, max } = arys.reduce( (acc, ary) => { const [start, end] = ary; let { min, max } = acc; if (start < min) { min = start; } if (end > max) { max = end; } return { min, max }; }, { min: Infinity, max: 0 } ); const hours = Array.from(generateAry(min, max)); const memo = hours.map(x => 0); function helper(arys, hours, memo) { let countRooms = 0; for (let row in arys) { for (let col in hours) { let prevCount = memo[col]; let [min, max] = arys[row]; // if the min > hour, set previous row same col value, the same as max < hour if (min > hours[col] || max < hours[col]) { memo[col] = prevCount; continue; } if (min <= hours[col] && max >= hours[col]) { memo[col] = prevCount + 1; } countRooms = Math.max(countRooms, memo[col]); } } console.log(memo); // [1, 3, 3, 2, 2, 2, 2, 1] return countRooms; } return helper(arys, hours, memo); } console.log(availableRoom([[2, 5], [1, 3], [2, 7], [6, 8]])); // 3